A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?

Step-1

What is given in problem?

In this problem mass of glucose is 10%w/w

So take mass of total solution = 100 g

Then mass of glucose will = 10% of 100 g = 10 grams

Mass of water will = 100 – 10 = 90 grams

And density = 1.2gmL-1

Step-2

find molality

Molality = number of moles of solute / mass of solvent in Kg   ……(1)

Here have to find number of moles and mass of solvent in Kg both

mass of solute  = mass of glucose / molar mass of glucose  …..(2)

Molar mass of glucose (C6H12O6)  = 6*C +  12*H + 6*O

= 6*12 + 12*1 +  6*16

=180 g /mol

Plug the value back in equation (2) we get

Number of moles =  10/180  = 0.0556 moles

We have mass of solvent (H2O)  =  90 g

Mass in Kg  = 90 g* 1 Kg / 1000 g    = 0.09 Kg

Plug the value in equation (1) we get

Molality  =  0.0556/0.09   = 16.7M

Step – 3 

Find molar fraction

Molar fraction of glucose  = number of moles of glucose / total number of moles

Molar mass of water (H2O) = 2*H+ O  = 2*1 + 16  = 18 g/ mol

Number of moles of water  = 90/18 = 5 moles

Plug the value in formula we get

 Mole fraction  = 0.0556/(0.0556+ 5)  = 0.01

It is binary solution because it has 2 component only so molar fraction of

Water  =1 – mole fraction of glucose

            = 1-0.01  =0.99

Step – 4

Find the molarity ,

molarity = number of moles of solute / volume of solution in liter        ……….(1)

Volume = mass / density

            = 100 g / 1.2 g/ ml  = 83.33 ml

Mass in liter  = 83.33 * 1 liter /1000 ml   = 0.0833liter

Plug the value back equation (1) we get

Molarity = 0.0556/ 0.0833

               = 0.67 M

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Answer :-

Step-1  find the given values

Mass % of nitric acid  = 68%

Density of nitric acid  = 1.504 g mL-1

Step -2

Let take total mass of solutions = 100    (why ?)

In this problem mass % is given whenever % is given take total quantity as 100 g ,

So  nitric acid =  68% of 100 g   = 68 g

Water will = 100 – 68  =  32 g

Step-3

We have to find molarity

Molarity(m) =  number of moles of solute / volume of solution in liter           ………..(1)

Here we have to find number of moles of solute  and volume of solution in liter                    

 Number of moles of solute = mass of solute /molar mass of solute             …………(2)

So molar mass of nitric acid HNO3  = 1*H + 1*N + 3*O  = 1*1 + 1*14 + 3*16  = 1+ 14 + 48 = 63 g

Plug the value in equation (2)  we get

Number of moles  = 68/63 = 1.079 moles

Volume = mass / density

              = 100g/1.504 g mL–1    =  66.5 ml

Now convert it in liter we get 

                                    66.5 ml * 1 liter / 1000ml =  0.0665 mL

Plug the value in equation first now we get

Molarity  = 1.079/ 0.0665  =16.23 M

Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What kind of solid solution is this likely to be?

Interstitial solid solution
In this solution, particles of solid solute fit into voids or interstices of solvent.

Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage

 

(i) Mole fraction (XA):-

It is the ratio of number of moles of one component to the total number of moles present in that solution

Let take a binary solution in which A is solute and B is solvent then

Molar fraction (XA) = number of moles of A/(number of moles of A + Number of moles of B)

If there are there are n number of component  in solution then

Molar fraction of component A =   number of moles of A / total number of moles

Sum of molar fraction of all components will always = 1

ie  NA+ NB+NC …………….= 1  (NA, NB  and NC .. are number of moles of components)

(ii) Molality(m) :-

Molality of any solution is represent the number of moles of solute present in per kg of solvent.

Molality(m) =  number of moles of solute / mass of solvent in Kg

Let take mass of solvent is = 1 kg then

Molality is numerically equal to number of moles of solute

So we can write the definition of molality that  “ molality  is the number of moles of solute present in 1 kg of solvent  “

Example 1.   Find the molality of solution containing 3 moles of  NaCl in 2 liter water ?

Density of water is 1kg/ liter so that    2 liter water = 2 kg of water

 Use above formula

    Molarity = 3/ 2 = 1.5 m

Calculate the molality of a solution containing 164 g of HCl in 753g of H20(water)

Number of moles of HCl = given mass of HCl / molar mass of HCl

Molar mass of HCl =  35.5 + 1  = 36.5 g /mol

Number of moles of HCl   = 164/36.5

                                           =4.5 moles

Mass of solvent in Kg  = 753 g* 1Kg/100 g  =  0.753 Kg

Molality = 4.5/0.753

            = 5.97m

Molarity(M):-

Molarity of any solution  is number of moles of solute per liter of solution

Molarity = number of moles of solute / volume of solution in liter

When volume of solution is 1 liter

Then molarity is numerically equal to number of moles of solute

So we can write a another definition of molarity that  “molarity is equal to number of moles of solute present in 1 liter of solution”

Mass percentage :-

(i) Mass percentage (w/w): The mass percentage of a component A of a solution is defined as mass of component A present in 100 unit mass of the solution. 

Mass % of a component A = mass of component A *100/Total mass of the solution

For example  let consider  a solution which has 15% salt in water by mass,  it has 15 g of glucose is dissolved in 85 g of water  in a 100 g solution.

(ii) Volume percentage (v/v): The volume percentage of a component A is defined as volume of component A present in 100 unit volume of the solution.

 Volume % of a component A = volume of component A *100/Total volume of the solution

For example, 15% methanol solution in water means that 15 mL of

Methanol is present in 100  ml of solution , and water will 100 – 15 = 85 ml

(iii) Mass by volume percentage (w/v): ): The mass by volume  percentage of a component A is defined as mass of  component A in gram present  100 ml volume of the solution.

Mass  by volume % of a component A = mass of component A in gram  *100/Total total volume  of the solution in ml .

solutions chemistry class 12 chapter 2

S.No.	Solution
1	Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
2	Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What kind of solid solution is this likely to be?
3	Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage
4	Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?
5	A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?
6	How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
7	A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
8	An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?
9	A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample.
10	What role does the molecular interaction play in a solution of alcohol and water?
11	Why do gases always tend to be less soluble in liquids as the temperature is raised?
12	State Henry’s law and mention some important applications?
13	The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?
14	What is meant by positive and negative deviations from Raoult's law and how is the sign of ΔmixH related to positive and negative deviations from Raoult's law?
15	An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
16	Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
17	The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
18	Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
19	A solution containing 30 g of non-volatile solute exactly in 90 g of water has a  vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
20	A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
21	Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B.
22	At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
23	Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water    (iv) methanol and acetone (v) acetonitrile (CH3CN) and acetone (C3H6O).
24	Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
25	Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.
26	If the density of some lake water is 1.25g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
27	If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
28	Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
29	Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 – 10–3 m aqueous solution required for the above dose.
30	Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
31	The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
32	Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1.
33	19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
34	Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
35	Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
36	100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
37	Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is: 100 x xacetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 pacetone /mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 pchloroform /mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
38	Benzene and toluene form ideal solution over the entire range of composition The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.
9	The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
40	Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.
41	Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated

S.No.	In text Questions
1	Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
2	Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
3	Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2.6H2O  in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
4	Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
5	Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
6	H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.
7	Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
8	The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
9	Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
10	Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol−1.
11	Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.
12	Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

ncert solutions for class 11 maths

S.No.	Ncert solutions for class 11 maths
1	Sets
2	Relations and Functions
3	Trigonometric Functions
4	Principle of Mathematical Induction
5	Complex Numbers and Quadratic Equations
6	Linear Inequalities
7	Permutations and Combinations
8	Binomial Theorem
9	Sequences and Series
10	Straight Lines
11	Conic Sections
12	Introduction to Three Dimensional Geometry
13	Limits and Derivatives
14	Mathematical Reasoning
15	Statistics
16	Probability

Welcome to NCERT Help

Welcome To NCERT HELP

We are trying to provide a new way of study, now  you can study and get higher marks without paying a single panny

ncerthelp.com is a website which offers you tutorials for your study and educational helps along with model test papers of each subject. It is the only website which provide education course and tips totally free. 

Read More

Mathematics-XII-Matrix-03-01-05(i)

If a matrix has 18 elements if it has 5 elements? what are the possible orders it can have? What

English

Hindi

If a matrix has 24 elements if it has 13 elements? what are the possible orders it can have? What

English

Hindi